### Problem Statement**Holding a Lead Ball with an Outstretched Arm**A person holds a 10.0 kg lead ball in their hand, at a distance of 32.0 cm from the elbow joint, as shown in the diagram. The biceps, attached at a distance of 2.5 cm from the elbow, exerts an upward force on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.50 kg. Calculate the force provided by the biceps muscle in order to hold the arm horizontally as shown. Use the elbow joint as the pivot point.### Diagram DescriptionThe diagram illustrates a forearm holding a ball. Key measurements indicated in the diagram are:- The ball is held 32 cm from the elbow joint.- The biceps muscle attaches to the forearm at a point 2.5 cm from the elbow joint.A detailed drawing shows the biceps in red and highlights the placement of forces and distances essential for solving the problem.### Solution ApproachTo solve this problem, we will employ the principles of torque (also known as moments). The torque created by the weight of the ball and the weight of the forearm must be balanced by the torque created by the biceps muscle's force. The elbow joint acts as the pivot point for calculating these torques.1. **Calculate the torques exerted by the weights:** - Torque due to the lead ball = \( (mass \; of \; ball) \times (gravitational \; acceleration) \times (distance \; from \; pivot) \) - Torque due to the forearm's weight should be considered at its center of mass, which is at half its length.2. **Calculate the torque exerted by the biceps muscle:** - Torque by biceps = \( (Force \; of \; biceps) \times (distance \; from \; pivot) \)3. **Set up the equilibrium condition for torques:** - Sum of clockwise torques = Sum of counterclockwise torques4. **Solve for the biceps muscle's force:**### Key Concepts- **Torque** is a measure of the force that can cause an object to rotate about an axis. It is given by the product of force and the distance from the pivot point (lever arm).- **Equilibrium** implies that the sum of

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